Leray-Serre spectral sequence for projective bundles

Let $$\mathcal{E} \rightarrow X$$ be a complex vector bundle of rank $$r+1$$ and let $$F=\mathbb{P}^r \rightarrow E = \mathbb{P}\mathcal{E}\rightarrow X$$ be the associated projective bundle. We know that under the assumption that $$H^{\bullet}(E) \rightarrow H^{\bullet}(F)$$ is surjective, by the Leray-Serre spectral sequence we have for any field $$k$$ $$H^{\bullet}(E, k) \simeq H^{\bullet}(X,k) \otimes H^{\bullet}(F,k)$$ Where the tensor product on the right is graduated. Can we say anything more about the algebra structure of left hand side when we take $$\mathbb{Z}$$ coefficients, maybe in terms of the Chern classes of the bundle, unravelling the definitions of the Leray-Serre spectral sequence?

As you say, if the map $$H(E)\to H(F)$$ is surjective, the Leray-Serre spectral sequence degenerates at the $$E^2$$-page which is therefore isomorphic to the $$E^\infty$$-page. Unwrapping the definitions, this means that there is a filtration on $$H^*(E)$$ such that the associated graded is $$H^*(X)\otimes H^*(F)\cong H^*(X)[x]/x^{r+1}$$. Choosing a representative of the class $$x$$, we obtain that $$H^*(E)$$ is generated as a $$H^*(X)$$-algebra (with $$H^*(X)$$ in filtration degree $$0$$) by a single element $$x$$ in degree and filtration degree $$2$$ such that $$x^{r+1}$$ sits in degree less than $$2r+2$$. This can only happen if there is a relationship of the form $$x^{r+1} = \sum_{k=0}^r p^*\gamma_{r+1-k}x^k$$ with $$\gamma_k\in H^{2k}(X)$$. Determining the concrete values of the $$c_k$$ amounts to solving the extension problem of the Leray-Serre spectral sequence, which is in a sense precisely the part of the problem the spectral sequence doesn't capture. However, in this case we can solve this problem geometrically: By construction, the pullback of $$\mathcal E$$ to $$E$$ splits off a line bundle $$L$$, and we may take $$x = -c_1(L)$$. Since $$p^*\mathcal E/L$$ is $$r$$-dimensional, its $$(r+1)$$-th Chern class vanishes, and the sum formula for the total Chern class yields $$c(p^*\mathcal E/L) = c(p^*\mathcal E)(1-x)^{-1}.$$ Expanding the second factor as $$\sum_{k=0}^r x^k$$ and taking the resulting expression for the $$(r+1)$$-th Chern class yields $$\gamma_k = c_k(\mathcal E)$$. In fact, this can be used to define Chern classes for general complex oriented cohomology theories.
• When you say: "The pullback splits off a line bundle" you mean you take $L$ to be line subbundle of $p^{\ast}\mathcal{E}$ generated by the tautological section? – Federico Feb 18 at 16:25
• More or less - a point in $E$ is a point $x\in X$, together with a one-dimensional subspace $V$ of $\mathcal E_x$. The fiber of $L$ over $(x,V)$ is precisely the one-dimensional subspace $V$ of $(p^*\mathcal E)_{(x,V)}\cong \mathcal E_x$. Note that $L$ is not generated by a non-vanishing section (otherwise it would be trivial, but by definition its restriction to every fiber is nontrivial), and there is no canonical section of $p^*\mathcal E$. – Bertram Arnold Feb 18 at 20:04